Let \(f, g: \mathbb{R} \to \mathbb{R}\) be the functions defined by \(f(x) = -x\) and \(g(x) = x + 1\). Let \(G\) be the subgroup generated by \(f\) and \(g\) of the group of all permutations of the set \(\mathbb{R}\). Prove carefully that \[G \cong \langle a, b \mid a^2 = e,\; ab = b^{-1}a \rangle.\]
Solution
Let \[ H=\langle a,b\mid a^2=e,\ ab=b^{-1}a\rangle. \]
1. The assignments\[ a\longmapsto f, \qquad b\longmapsto g \]define a surjective homomorphism \(\Phi:H\to G\).
The generators \(f,g\) satisfy the defining relations of \(H\). First, \[ f^2(x)=x, \] so \(f^2=e\). Also \[ (fg)(x)=f(x+1)=-x-1 \] and \[ (g^{-1}f)(x)=g^{-1}(-x)=-x-1. \] Hence \[ fg=g^{-1}f. \] Therefore the presentation of \(H\) gives a homomorphism \[ \Phi:H\to G \] with \(\Phi(a)=f\) and \(\Phi(b)=g\). Since \(G=\langle f,g\rangle\), this homomorphism is surjective.
2. Every element of \(H\) has the form\[ b^n \qquad\text{or}\qquad b^n a \]for some \(n\in\mathbb Z\).
From \[ ab=b^{-1}a \] we obtain, by induction, \[ ab^n=b^{-n}a \qquad(n\in\mathbb Z). \] Thus any occurrence of \(a\) can be moved to the far right of a word, at the cost of inverting the exponent of the adjacent power of \(b\). After doing this throughout the word, all copies of \(a\) occur at the far right. The relation \(a^2=e\) then reduces their number modulo \(2\). Hence every element has one of the stated forms.
3. The homomorphism \(\Phi\) is injective.
For \(n\in\mathbb Z\), \[ \Phi(b^n)=g^n, \qquad \Phi(b^n a)=g^n f. \] These transformations are \[ g^n(x)=x+n, \qquad (g^n f)(x)=-x+n. \] If \(g^n\) is the identity permutation of \(\mathbb R\), then evaluating at \(0\) gives \(n=0\). On the other hand, \(g^n f\) is never the identity, since it has slope \(-1\); explicitly, equality \(-x+n=x\) for every \(x\) is impossible.
By 2, every element of \(\ker\Phi\) is either \(b^n\) or \(b^n a\). The preceding calculation shows that only \(b^0=e\) lies in the kernel. Therefore \[ \ker\Phi=\{e\}. \]
4. Hence\[ G\cong \langle a,b\mid a^2=e,\ ab=b^{-1}a\rangle. \]
By 1, \(\Phi\) is surjective, and by 3 it is injective. Thus \(\Phi\) is an isomorphism.